Question: Each of $a_1,$ $a_2,$ $\dots,$ $a_{100}$ is equal to $1$ or $-1.$  Find the minimum positive value of
\[\sum_{1 \le i < j \le 100} a_i a_j.\]
Solution: Let $S$ denote the given sum.  Then
\begin{align*}
2S &= (a_1 + a_2 + \dots + a_{100})^2 - (a_1^2 + a_2^2 + \dots + a_{100}^2) \\
&= (a_1 + a_2 + \dots + a_{100})^2 - 100.
\end{align*}To find the minimum positive value of $2S,$ we want $(a_1 + a_2 + \dots + a_{100})^2$ to be as close to 100 as possible (while being greater than 100).  Since each $a_i$ is $1$ or $-1,$ $a_1 + a_2 + \dots + a_{100}$ must be an even integer.  Thus, the smallest we could make $(a_1 + a_2 + \dots + a_{100})^2$ is $12^2 = 144.$  This is achievable by setting 56 of the $a_i$ to be equal to $1,$ and the remaining 44 to be equal to $-1.$

Thus, the minimum positive value of $S$ is $\frac{144 - 100}{2} = \boxed{22}.$